Institut für Analysis WS2019/20 Prof. Dr. Dorothee Frey 28.11.2019 M.Sc. Bas Nieraeth
Functional Analysis
Solutions to exercise sheet 7
Exercise 1: Strong convergence on compact sets. Let X and Y be Banach spaces. Suppose there is sequence of operators (Sn)n∈N in L(Y ) such that Sn converges strongly to the identity operator on Y , i.e., for all y ∈ Y we have
lim kSny − yk = 0. n→∞
(a) Show that for each compact set K ⊆ Y we have
lim sup kSny − yk = 0. n→∞ y∈K
(Hint: Find a suitable open cover of K. Also use the Banach-Steinhaus Theorem.)
(b) Show that for each compact operator T ∈ K(X,Y ) we have that SnT → T as n → ∞ with respect to the operator norm.
(c) Let p ∈ [1, ∞). Show that any compact operator T ∈ K(X, `p) is the operator norm limit of a sequence of finite rank operators.
Solution:
(a) First note that since for each y ∈ Y the sequence (Sny)n∈N is convergent, this sequence is also bounded. Thus, by the Uniform Boundedness Principle we have supn∈N kTnk < ∞.
ε Let ε > 0 and set ε˜ := sup kS k+2 . By our assumption on the Sn, for each y ∈ Y we can n∈N n find an Ny ∈ N such that kSny − yk < ε˜ for n ≥ Ny. (1) Since K is compact, it is totally bounded. Thus, we can find a finite subset F ⊆ K such that (B(˜y, ε˜))y˜∈F covers K. Now set N := maxy˜∈F Ny˜.
Let y ∈ K and pick y˜ ∈ F such that y ∈ B(˜y, ε˜). Then if n ≥ N, we have n ≥ Ny˜ so that by (1) we have
kSny − yk ≤ kSny − Sny˜k + kSny˜ − y˜k + ky˜ − yk
< kSnkky − y˜k +ε ˜+ε ˜
< (sup kSnk + 2)˜ε = ε. n∈N Thus, whenever n ≥ N we have
sup kSny − yk ≤ ε, y∈K proving the assertion. — Turn the page! — (b) As T is compact, the set T (BX ) is compact. Hence, by part (a),
kSnT − T k = sup kSnT x − T xk = sup kSny − yk ≤ sup kSny − yk → 0 kxk≤1 y∈T (Bx) y∈T (Bx) as n → ∞.
p p (c) We define Sn : ` → ` as the projection on the first n coordinates, i.e., Sny := (y1, . . . , yn, 0, 0,...).
Then 1 ∞ p X p kSny − ykp = |yj| → 0 as n → ∞. j=n+1 p Now let T ∈ K(X, ` ). Since SnT maps into the range of Sn, which is finite dimensional, SnT is a finite rank operator. Moreover, by part (b) we have limn→∞ kSnT − T k = 0. We conclude that T is the operator norm limit of finite rank operators, as desired.
p Exercise 2: Translations in L (R).
(a) Let X be a Banach space and suppose we have a family of operators (T (t))t∈R in L(X) satisfying the properties that T (s + t) = T (s)T (t) for all s, t ∈ R and T (0) = Id. Show that the following are equivalent.
(i) For each x ∈ X the map R → X, t 7→ T (t)x is continuous.
(ii) limt→0 T (t)x = x in X for all x ∈ X;
(iii) For every sequence (tn)n∈N with tn → 0 we have supn∈N kT (tn)k < ∞. Moreover, there is a dense subset D ⊆ X such that limt→0 T (t)x = x in X for all x ∈ D.
p p (b) Let p ∈ [1, ∞). For each t ∈ R we define T (t): L (R) → L (R) by (T (t)f)(x) = f(x + t). Show that the map p R → L (R), t 7→ T (t)f
is continuous. You may use without proof that the space Cc(R) of continuous functions on R of compact support (i.e., the set of continuous functions f : R → R such that p {x ∈ R : f(x) 6= 0} is bounded) is dense in L (R).
Solution:
(a) We will prove (i)⇔(ii) and (ii)⇔(iii).
“(i)⇒(ii)”
Since the map R → X, t 7→ T (t)x is, in particular, continuous at 0, we have, since T (0) = Id, that limt→0 T (t)x = T (0)x = x for all x ∈ X. This proves (ii).
“(ii)⇒(i)”
Fix t0 ∈ R. We will show that R → X, t 7→ T (t)x is continuous at t0. For x ∈ X we have
lim T (t)x = lim T (t − t0)T (t0)x = lim T (s)T (t0)x = T (t0)x t→t0 t→t0 s→0 by (ii). Thus, the map R → X, t 7→ T (t)x is continuous at every point, and hence, is continuous.
“(ii)⇒(iii)”
Let (tn)n∈N be a sequence converging to 0. Then by (ii) we have limn→∞ T (tn)x = x for every x ∈ X. In particular, the sequence (T (tn)x)n∈N is bounded for each n ∈ N. Hence, by the Uniform Boundedness Principle, we have supn∈N kT (tn)k < ∞.
For the second assertion in (iii), we can set D = X.
“(iii)⇒(ii)”
Let x ∈ X. Note that the assertion limt→0 T (t)x = x is equivalent to the assertion that for every sequence (tn)n∈N converging to 0, we have limn→∞ T (tn)x = x. We prove this latter assertion.
ε Let (tn)n∈N be a sequence with tn → 0 and let ε > 0. We set ε˜ = sup kT (t )k+2 and pick n∈N n y ∈ D such that kx − yk < ε˜ and N ∈ N such that for all n ≥ N we have kT (tn)y − yk < ε˜. Then for all n ≥ N we have
kT (tn)x − xk ≤ kT (tn)x − T (tn)yk + kT (tn)y − yk + ky − xk
< kT (tn)kkx − yk +ε ˜+ε ˜
< (sup kT (tn)k + 2)˜ε = ε. n∈N
Thus, limn→∞ T (tn)x = x and the assertion follows.
(b) First note that T (s + t) = T (s)T (t) for all s, t ∈ R and T (0) = Id. Thus, to conclude p the result, it suffices to prove (iii) of part (a). First note that forall f ∈ L (R) and t ∈ R we have kT (t)fkp = kfkp by translation invariance of the Lebesgue measure. Hence, we have kT (t)k = 1 for all t ∈ R and thus, in particular, for any sequence tn → 0, supn∈N kT (tn)k = 1. It remains to check the second condition of (iii).
We set D := Cc(R). Let f ∈ D and let r > 0 so that {x ∈ R : f(x) 6= 0} ⊆ [−r, r]. We claim that f is uniformly continuous. For now, we assume the claim.
Let ε > 0. By uniform continuity of f we can pick δ˜ > 0 such that if |t| < δ˜, then p εp ˜ |f(x + t) − f(x)| < 2r+2 for all x ∈ R. Then set δ := min{1, δ}.
If |t| < δ, then |t| < δ˜ and {x ∈ R : f(x + t) 6= 0} ⊆ [−r − 1, r + 1]. Hence, Z Z r+1 p p p p ε p kT (t)f − fkp = |f(x + t) − f(x)| dx = |f(x + t) − f(x)| dx ≤ (2r + 2) = ε R −r−1 2r + 2 p so that kT (t)f − fkp ≤ ε. Hence limt→0 T (t)f = f in L (R). The assertion follows.
Finally, we prove the claim. Note that since the set [−r − 1, r + 1] is compact, the function f restricted to this set is uniformly continuous. Thus, for ε > 0 we can pick δ˜ > 0 such that whenever x, y ∈ [−r − 1, r + 1] satisfy |x − y| < δ˜, then |f(x) − f(y)| < ε.
Set δ := min{1, δ} and let x, y ∈ R such that |x − y| < δ. We consider two cases. In the first case, we assume that x, y ∈ [−r − 1, r + 1]. Then, since |x − y| < δ ≤ δ˜, we have |f(x) − f(y)| < ε, as desired. In the other case we have x∈ / [−r − 1, r + 1] or — Turn the page! — y∈ / [−r − 1, r + 1]. We assume without loss of generality that x∈ / [−r − 1, r + 1]. But then, since |x − y| < δ ≤ 1, we have y∈ / [−r, r]. This means that f(x) = f(y) = 0 so that |f(x) − f(y)| = 0 < ε, as desired. We conclude that f is indeed uniformly continuous, proving the claim.
Exercise 3: Duality in sequence spaces. Let y = (yj)j∈N be a sequence in K.
(a) Suppose that ∞ X xjyj j=1 1 converges in K for all (xj)j∈N ∈ c0. Show that y ∈ ` .
(Hint: use the Banach-Steinhaus Theorem.)
(b) Let p ∈ [1, ∞]. Suppose that ∞ X xjyj j=1
p p0 1 1 converges in K for all (xj)j∈N ∈ ` . Show that y ∈ ` , where p + p0 = 1.
Solution:
Pn (a) For x ∈ c0 we define Tnx := j=1 xjyj. Then Tn is linear and we have
n n X X |Tnx| ≤ |xjyj| ≤ kxk∞ |yj|, j=1 j=1
showing that Tn ∈ L(c0, K). Moreover, we assumed that (Tnx)n∈N is a convergent sequence in K for all x ∈ c0. Thus, by the Banach-Steinhaus Theorem we have supn∈N kTnk < ∞.
−1 Fix J ∈ N. we define a sequence x ∈ c0 by xj = yj|yj| (with the understanding that xj = 0 if yj = 0) for j ∈ {1,...,J} and xj = 0 for j > J. Then, kxk∞ ≤ 1 and
J X |yj| = |TJ x| ≤ kTJ k ≤ sup kTnk j=1 n∈N
1 Taking a supremum over J ∈ N proves that y ∈ ` , as desired.
p Pn (b) Then for x ∈ ` we define Tnx := j=1 xjyj and note that by Hölder’s inequality we have
1 n n p0 X X p0 |Tnx| ≤ |xjyj| ≤ kxkp |yj| . j=1 j=1
p Thus, Tn ∈ L(` , K). Since we assumed that the sequence (Tnx)n∈N is convergent for every p x ∈ ` , it follows from the Banach-Steinhaus Theorem that supn∈N kTnk < ∞.
We consider the cases p > 1 and p = 1 separately. First suppose p ∈ (1, ∞] (where in 1 0 p the case that p = ∞ we interpret p as 0). Then p ∈ [1, ∞) and we define x ∈ ` by p0−2 1 1 p0−1 xj = yj|yj| for j ∈ {1,...,J} and xj = 0 for j > J. Then, since p = 1 − p0 = p0 , we 0 0 0 p 0 p p −1 p have p − 1 = p so that |xj| = |yj| = |yj| for j ∈ {1,...,J}. Hence,
1 J p X p0 kxkp = |yj| j=1
and 1 1 1 J p0 J p J J p X p0 X p0 X p0 X p0 |yj| |yj| = |yj| = |TJ x| ≤ kTJ kkxkp ≤ sup kTnk |yj| . j=1 j=1 j=1 n∈N j=1
So that 1 J p0 X p0 |yj| ≤ sup kTnk. j=1 n∈N
p0 Taking a supremum over J ∈ N shows that y ∈ ` , as desired.
Now suppose p = 1. For j ∈ N, let ej denote the sequence that is 0 in every place except for the j-th place, where it is equal to 1. Then for each j ∈ N we have
|yj| = |Tjej| ≤ kTjk ≤ sup kTnk. n∈N ∞ Thus, taking a supremum over j ∈ N shows that y ∈ ` , as desired.
Exercise 4: Reverse estimates. Let X and Y be Banach spaces and T ∈ L(X,Y ).
(a) Show that T is injective and has a closed range R(T ) if and only if there is a c > 0 such that kT xk ≥ ckxk for all x ∈ X.
(Hint: use the Open Mapping Theorem).
(b) Show that T has a closed range R(T ) if and only if there is a c > 0 such that
kT xk ≥ c dist(x, N(T )) for all x ∈ X.
0 Here dist(x, N(T )) = infx0∈N(T ) kx − x k.
Solution:
(a) In Exercise 4 of Exercise sheet 4 we have shown that if the reverse estimate holds, then T is injective and R(T ) is closed. It remains to prove the converse result.
If R(T ) is closed in the Banach space X, then it is itself a Banach space. We define a new operator T˜ : X → R(T ) by T˜ x := T x. Then this map is surjective per construction. Moreover, it is injective, since T is injective. We conclude that T˜ is a bijective map and we denote its inverse by S : R(T ) → X.
Since T˜ is in particular surjective, it follows from the Open Mapping Theorem that for any open set U ⊆ X we have that T˜(U) is open in X. Hence, if U ⊆ X is open, then — Turn the page! — S−1(U) = T˜(U) is open in R(T˜) = R(T ), proving that S is a continuous and thus bounded operator. Thus, for all x ∈ X we have
kxk = kST xk ≤ kSkkT xk.
Setting c := (kSk + 1)−1 > 0 proves the assertion.
(b) Since T is bounded, N(T ) = T −1({0}) is closed. Thus, the quotient space X/N(T ) is again a Banach space. We define T˜ : X/N(T ) → Y by T˜xˆ := T x. Note that this is well-defined, since if x − y ∈ N(T ), then T x − T y = T (x − y) = 0 so that T x = T y.
Next, we note that T˜ is injective. Indeed, if T˜xˆ = 0, then T x = 0. Thus, x − 0 = x ∈ N(T ) so that xˆ = 0ˆ = 0, as desired.
Finally, we note that R(T˜) = R(T ). Thus, it follows from part (a) that R(T ) is closed if and only if there is a c > 0 such that
kT xk = kT˜xˆk ≥ ckx˜k = c dist(x, N(T )).
This proves the assertion.
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